Question

A child running a temperature of 101°F is given an antipyrin (i.e. a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98 °F in 20 min, what is the average rate of extra evaporation caused, by the drug? Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is 30 kg. The specific heat of human body is approximately the same as that of water, and latent heat of evaporation of water at that temperature is about 580 cal g–1.

Answer 1

Here,
mass of the child (m) = 30 kg
time taken ( T) = 20 min
Fall in temperature = (101-98)°F
= 3°F = 3 × 5/9 = 5/3 °C [ 1°F = 5/9° C ]
specific heat of human body ( S) = 4.2 × 10³ J/Kg-°C
Latent heat of vaporisation ( Lv) = 580 Cal/g = 580 × 10^3 Cal/kg
= 580 × 4.2 × 10³ j/kg

Use formula ,
Heat given by body (Q1) = mS ∆T
Let m' is the mass of sweat evaporates from the human body ,
Heat taken ( Q2) = m'Lv

Form calorimetry ,
Q1 = Q2
mS ∆T = m'Lv
m' = mS∆T/Lv
= 30 × 4.2 × 10³ × 5/3 /58 × 4.2 × 10³
= 10/116 Kg
= 0.0862 Kg